A typical nuclear reactor generates 1000 MW (1000 MJ/s) of electric energy. In doing so, it produces 2000 MW of 'waste heat' that must be removed from the reactor to keep it from melting down. Many reactors are sited next to large bodies of water so that they can use the water for cooling. Consider a reactor where the intake water is at 18°C. State regulations limit the temperature of the output water to 30°C so as not to harm aquatic organisms. How many liters of cooling water have to be pumped through the reactor each minute?

Question

Accepted Answer

Hey, everyone. So this problem is dealing with heat transfer. Let's see what it's asking us. We have a geothermal power plant that generates electricity during the process. 4110 megawatts of electric heat needs to be dissipated to cool the system. We, they pump water from a nearby river. The river has an initial temperature of 20 degrees Celsius and the return temperature cannot exceed 35 degrees Celsius. So we're asked to calculate the flow rate of the water in liters per minute required to maintain this amount of heat dissipation. Our multiple choice answers again in units of liter per minute are a 3.29, sorry 3.92 times 10 to the six B, 4.94 times 10 to the five C 3.31 times 10 to the five or D 4.45 times 10 to the six. So the first step to solve this problem is recall that our heat transfer equation is given by Q equals MC delta T where Q is given by the equation power multiplied by time and mass is given by the equation density multiplied by volume the mass equation can be further expanded based on the variables that we were given. And what we were asked for which is the flow rate to be the density multiplied by the volumetric flow rate multiplied by time. When we plug each of these expanded equations into our original equation, we're left with P T equals row, the sorry uh row multiplied by the volumetric flow rate or B dot multiplied by T multiplied by C multiplied by delta uppercase T. We're solving for this volumetric flow rate. So we can simplify the equation and isolate that variable. And we're left with P are power divided by row C delta T. And when we look at the problem, we're actually given each of these values. And so from here, it's just a plug and shot to find our final answer. So our power was given to us as 4110 megawatts as the amount of our electric energy that we need to dissipate. We want to keep everything in standard units. So we can rewrite that as 4. times 10 to the nine and we called it watts is the same as a jewel or set. And so all of that divided by our density density of water we can recall is one kg per liter. The specific heat of water, sorry specific heat capacity of water is 4190 jewels per kilogram Celsius. And then our delta T where our t final, that maximum temperature of 35 degrees C minus our T initial initial river temperature of 20 degrees Celsius. When we plug that into our calculator, we get 6.54 times 10 to the four liters per second. And now the last step is to recognize that the answers are given in leaders per minute. And so we can multiply this by 60 seconds for a minute. So using that conversion factor, we get 3.92 times to the sixth leaders are my. And that is the final answer to this problem and it aligns with answer choice. A so A is the correct choice here. That's all we have for this one. We'll see you in the next video.

Verified Solution

Key Concepts

Heat Transfer

Specific Heat Capacity

Flow Rate